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Question

Chemistry Question on Equilibrium

For the reaction AB(g)A(g)+B(g),ABA B(g) \rightleftharpoons A(g)+B(g), A B is 33%33 \% dissociated at a total pressure of pp. Therefore, pp is related to KpK_{p} by one of the following options

A

P=KpP = K_{p}

B

P=3KpP = 3K_{p}

C

P=4KpP = 4K_{p}

D

P=8KpP = 8K_{p}

Answer

P=8KpP = 8K_{p}

Explanation

Solution

Total pressure at equilibrium,
=10.33+0.33+0.33=1-0.33+0.33+0.33
=1+0.33=1.33=1+0.33=1.33
pA=0.331.33pp_{A} =\frac{0.33}{1.33} p
pB=0.331.33pp_{B} =\frac{0.33}{1.33} p
pAB=0.671.33pp_{A B} =\frac{0.67}{1.33} p
(Where, pA,pBp_{A}, p_{B} and pABp_{A B} are partial pressures of A,BA, B and ABA B.)
Kp=pApBpABK_{p} =\frac{p_{A} \cdot p_{B}}{p_{A B}}
=0.331.33p0.331.33p0.671.33p=\frac{\frac{0.33}{1.33} p \cdot \frac{0.33}{1.33} p}{\frac{0.67}{1.33} p}
=(0.33)2p1.33×0.67=\frac{(0.33)^{2} p}{1.33 \times 0.67}
=0.1089×p0.8911=\frac{0.1089 \times p}{0.8911}
Kp=0.122pK_{p} =0.122\, p
p=Kp0.122=8Kpp =\frac{K_{p}}{0.122}=8\, K_{p}